Intermediate Algebra, Chapter 2, 2.7, Section 2.7, Problem 72

Solve the inequality $|6x - 1| - 2 > 6$.


$
\begin{equation}
\begin{aligned}

|6x - 1| - 2 > & 6 \\
|6x - 1| > & 8

\end{aligned}
\end{equation}
$


The absolute value inequality is rewritten as

$6x - 1 > 8$ or $6x - 1 < -8$

because $6x - 1$ must represent a number that is more than $8$ units from on either side of the number line. We can solve the compound inequality.


$
\begin{equation}
\begin{aligned}

6x - 1 > & 8 && \text{or} &&& 6x - 1 < & -8
&&
\\
6x > & 9 && \text{or} &&& 6x < & -7
&& \text{Add } 1
\\
x > & \frac{9}{6} && \text{or} &&& x < & - \frac{7}{6}
&& \text{Divide each side by } 6
\\
\\
x > & \frac{3}{2} && \text{or} &&& x < & - \frac{7}{6}
&& \text{Reduce to lowest term}

\end{aligned}
\end{equation}
$


The solution set is $\displaystyle \left( - \infty, - \frac{7}{6} \right) \bigcup \left( \frac{3}{2}, \infty \right)$.